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Econometrics Regression Problem Set Due date: 2/8/10 Problems 1) Problem 4.2 in Stock and Watson (SW) (Note: 1 lb = 2.54 in and 1 lb = .4536 kg) 2) Problem 4.5 in SW 3) Problem 4.6 in SW 4) Problem 4.7 in SW 5) Suppose that you run the regression: Yi 0 1 X i ui Using ordinary least squares, you get the following estimates of the regression coefficients: ˆ1 ( X i X )(Yi Y ) ( X i X )2 and ˆ0 Y ˆ1 X Now suppose you define a new variable Zi = 2Xi and you run the regression: Yi 0 1Zi ei Using ordinary least squares, you get the following estimates of the new regression coefficients: ˆ1 ( Zi Z )(Yi Y ) (Z 2 i Z) and ˆ 0 Y ˆ1Z Your job: Express ̂ 0 and ̂1 as a function of ̂ 0 and ̂1 . 6) Problem 5.3 in Stock and Watson (SW) 7) Problem 5.4 in SW 8) Problem 5.5 in SW (Assume that the sample size is over 1000) 9) Problem 5.10 in SW (Hint: Let n0 denote the number of observation with X 0 and n1 denote the n number of observations with X 1; note that i 1 X i n1 ; ) Also, write a sentence to explain why it makes intuitive sense that ˆ1 Y1 Y0 . 1. (a) Substituting Height 70, 65, and 74 inches into the equation, the predicted weights are 176.39, 156.69, and 192.15 pounds. (b) Weight 394 Height 394 15 591. (c) We have the following relations: 1 in 254 cm and 1 lb 04536 kg. Suppose the regression equation in the centimeter-kilogram space is Weight ˆ 0 ˆ1 Height . The coefficients are ˆ 0 9941 04536 45092 kg; ˆ1 394 024536 07036 kg per 54 cm. The R2 is unit free, so it remains at R2 081 . 2. (a) ui represents factors other than time that influence the student’s performance on the exam including amount of time studying, aptitude for the material, and so forth. Some students will have studied more than average, other less; some students will have higher than average aptitude for the subject, others lower, and so forth. (b) Because of random assignment ui is independent of Xi. Since ui represents deviations from average E(ui) 0. Because u and X are independent E(ui|Xi) E(ui) 0. (c) (2) is satisfied if this year’s class is typical of other classes, that is, students in this year’s class can be viewed as random draws from the population of students that enroll in the class. (3) is satisfied because 0 Yi 100 and Xi can take on only two values (90 and 120). (d) (i) 49 0.24 90 70.6; 49 0.24 120 77.8; 49 0.24 150 85.0 (ii) 0.24 10 2.4. 3. Using E(ui |Xi ) 0, we have E(Yi |Xi ) E(0 1Xi ui |Xi ) 0 1E( Xi |Xi ) E(ui |Xi ) 0 1Xi 4. The expectation of ˆ 0 is obtained by taking expectations of both sides of Equation (4.8): 1 n E ( ˆ0 ) E (Y ˆ1 X ) E 0 1 X ui ˆ1 X n i1 n 1 0 E ( 1 ˆ1 ) X E (ui |Xi ) 0 n i1 where the third equality in the above equation has used the facts that ̂1 is unbiased so E( ˆ ) 0 and E(u |X ) 0. 1 1 i i 5. ˆ1 ( Zi Z )(Yi Y ) (Z i Z) 2 = (2 X 2 X )(Y Y ) 2 ( X X )(Y Y ) 1 ˆ 4 ( X X ) 2 (2 X 2 X ) i i i i 2 i 2 i 1 So: 1 2 ˆ 0 Y ˆ1Z Y ˆ1 (2 X ) Y ˆ1 (2 X ) Y ˆ1 X ˆ0 We find that ̂1 1 ˆ 1 and ̂0 ˆ0 . 2 6. The 99% confidence interval is 1.5 {3.94 2.58 0.31) or 4.71 lbs WeightGain 7.11 lbs. 7. (a) 3.13 1.47 16 $20.39 per hour (b) The wage is expected to increase from $14.51 to $17.45 or by $2.94 per hour. (c) The increase in wages for college education is 1 4. Thus, the counselor’s assertion is that 1 10/4 2.50. The t-statistic for this null hypothesis is 2.50 t 1.470.07 14.71, which has a p-value of 0.00. Thus, the counselor’s assertion can be rejected at the 1% significance level. A 95% confidence for 1 4 is 4 (1.47 1.97 0.07) or $5.33 Gain $6.43. 8. (a) The estimated gain from being in a small class is 13.9 points. This is equal to approximately 1/5 of the standard deviation in test scores, a moderate increase. 5.56, which has a p-value of 0.00. Thus the null (b) The t-statistic is t act 13.9 2.5 hypothesis is rejected at the 5% (and 1%) level. (c) 13.9 2.58 2.5 13.9 6.45. 9. Let n0 denote the number of observation with X 0 and n1 denote the number of observations with n n X 1; note that i 1 X i n1 ; X n1|n; n1 i 1 X iYi Y1 ; 1 n ( X i X ) i 1 X nX n1 n 2 i 1 n Y n1 Y1 n0 n 2 i 2 n12 n n1 1 n1 n n1n0 n ; n1Y1 n0Y0 i 1Yi , so that Y0 From the least squares formula ˆ1 in1 ( Xi X )(Yi Y ) in1 Xi (Yi Y ) in1 XiYi Yn1 n in1 ( Xi X )2 i1 ( Xi X )2 n1n0 |n n n n n (Y1 Y ) Y 1 Y1 0 Y0 Y1 Y0 n0 n0 n n n n n n0 n Y0 Y0 and ˆ0 Y ˆ1 X 0 Y0 1 Y1 (Y1 Y0 ) 1 1 n n n n n